If `w=alpha+ibeta,` where `beta!=0` and `z!=1` , satisfies the condition that `((w- bar w z)/(1-z))` is a purely real, then the set of values of `z` is `|z|=1,z!=2` (b) `|z|=1a n dz!=1` (c)`z=bar z ` (d) None of these

Options

A. `{z{|z|=1}`
B. `{z:z=barz}`
C. `{z:z ne 1}`
D. `{z:|z|=1, z ne 1}`

Answer

D

Explanation

It is given that `(omega-baromegaz)/(1-z)` is purely real.
`therefore (omega-bar(omega)z)/(1-z)=(bar(omega-baromegaz))/(1-z)`
`rArr baromega-baromegaz-omegabarz+baromegazbarz=baromega-omegabarz-baromegaz-baromegaz+omegazbarz`
`baromega-omega+omega|z|^(2)-baromega|z|^(2)=0`
`rArr (baromega-omega)(1-|z|^(2))=0`
`rArr -2ibeta(1-|z|^(2))=0 rArr |z|=1 [therefore beta ne 0]`
Also, `z ne 1`
Hence, the set of values of z is `{z:|z|=1, z in 1}`.